Given two symmetric p.d. matrices of dimension $(p \times p)$, $\mathbf{A}$ and $\mathbf{B}$, can we interpret $$ d = \operatorname{trace}(\mathbf{A}^{-1} \mathbf{B}) $$ as a distance between the two?
For this it would obviously have to be the case that $d$ is minimized if $\mathbf{A} = \mathbf{B} \Rightarrow d = p$. Is this the case?
The answer is no. Consider the case $A,B$ are diagonal positive definite matrices. It is easy to make $d$ arbitrarily small/arbitrarily close/equal to $p$ without the matrices being close/equal.
To make this clear, let $x_1,..., x_p$ be arbitrary positive numbers with $x_1+...+x_p=p$. Set $A=I, B= \mbox{diag}(x_1,...,x_p)$.
Then $d(A,B)=p$.